$\int e^{2x} (\sin x +\frac{1}{2} \cos x) dx$ is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{2}e^{2x}\sin x+C$, C is an arbitrary constant
$I = \int e^{2x}\left(\sin x + \frac{1}{2}\cos x\right)dx$
$I = \int e^{2x}\sin x\,dx + \frac{1}{2}\int e^{2x}\cos x\,dx$
Using standard results:
$\int e^{ax}\sin bx\,dx = \frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx - b\cos bx)$
$\int e^{ax}\cos bx\,dx = \frac{e^{ax}}{a^{2}+b^{2}}(a\cos bx + b\sin bx)$
Substitute $a=2,\;b=1$:
$I = \frac{e^{2x}}{5}(2\sin x - \cos x) + \frac{1}{2}\cdot\frac{e^{2x}}{5}(2\cos x + \sin x) + C$
$I = \frac{e^{2x}}{10}\left[(4\sin x - 2\cos x) + (2\cos x + \sin x)\right] + C$
$I = \frac{e^{2x}}{10}(5\sin x) + C$
$I = \frac{1}{2}e^{2x}\sin x + C$